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${C_P}\;and\,\;{C_v}$are specific heats at constant pressure and constant volume respectively. It is observed that ${C_P}\, - \,{C_V}\, = \,a$ for hydrogen gas ${C_P}\, - \,{C_V}\, = \,b$for nitrogen gas.The correct relation between a and b is
A. $a\, = \,28b$
B. $a\, = \,\dfrac{1}{{14}}b$
C. $a\, = \,b$
D. $a\, = \,14b$

Answer
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Hint: Concept of conductivity and the factors on which it depends. Good conductivity refers to a good conductor.

Complete step by step answer:
By Mayer’s formula, for 1 g mole of a gas
${C_P} - {C_V} = R$
Where ${C_P}$ is the specific heat at constant pressure
${C_V}$ is the specific heat at constant volume and R is the gas constant.
Now, when gram moles are given, this formula reduces to
${C_P} - {C_V} = \dfrac{R}{M}$
Where
M is the molar mass
For Hydrogen gas, $M = 2 \times 1 = 2$
So, ${C_P} - {C_V} = \dfrac{R}{2}$ ….(1)
But according to question, ${C_P}\, - \,{C_V}\, = \,a$ …..(2)
$
  a = \dfrac{R}{2} \\
  R = 2a \\
 $ ….(3)
For Nitrogen gas, Molar mass, $M = 2 \times 14 = 28$
So, Mayer’s formula becomes
${C_P} - {C_V} = \dfrac{R}{{28}}$ ….(4)
But ${C_P}\, - \,{C_V}\, = \,b$ ….(5) (given)
From (4) and (5),
$b = \dfrac{R}{{28}}$ ….(6)
Putting equation (3) in (6), we get
$b = \dfrac{a}{{14}}$
$ \Rightarrow \;\;\;a = 14\,b$
So, D is the correct option.

Note: ${C_P}$ is greater than ${C_V}$ because when a gas is heated at constant volume, no external work is done and so the heat supplied is consumed only in increasing the internal energy of a gas. But if the gas is heated at constant pressure, the gas expands against the external pressure so does some external work, in this case the heat is used up in increasing the internal energy of the gas and in doing some external work. Since the internal energy depends only on temperature the internal energy of a mass of a gas will increase by the same amount whether the pressure or volume remains constant. But since external work is additionally done for constant pressure than at constant volume to produce the same rise in temperature of the gas.